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A is a solution containing 12.0g/dm3 of NaHSO4; b is a solution of NaOHrn25cm3 of solution b requires an average of 21.60cm3 of solution a for complete neutralization using methyl orange as indicator. The equation for the reaction involved in the titration is:rn NaHSO4(aq) + NaOH(aq) → Na2SO4(aq) + H2O(l) Calculatern(C) Mass of Na+ formed in the solution during the titration
Accepted Answer
C) Mass of Na+ formed in the solution during the titration
Principle:
1 mole of NaHSO4 reacts with 1 mole of NaOH to produce 1 mole of Na2SO4.
Stoichiometric Calculations
Moles of NaHSO4 = (12 g/dm3) x (25 cm3 x 10^-3 dm3 /25 cm3) = 0.0003
From the balanced equation,
1 mole of NaHSO4 produces 1 mole of Na+ ions
Therefore, 0.0003 moles of NaHSO4 will produce 0.0003 moles of Na+ ions.
Conversion to mass:
Mass of Na+ formed = (0.0003 mol) x (23 g/mol) = 0.0069 g
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